Hold your horses. ln(x)dx = 1/x is a direct result of e^xdx = e^x. You can't prove the rule with itself.

Also, (for some obscure reason) you've used the chain rule without proving its validity. That's no proof at all. It's just restating what you were supposed to prove.

I'll prove ln(x)dx = 1/x, too then. And what do you mean 'without proving its validity'? The chain rule has already been proven to work with any function, hasn't it?

e = lim (x->Math.huge) [(1 + 1/x)^x], by definition

dx[ln(x)] = lim (Δx->0) [(ln(x + Δx) - ln(x)) / Δx], by definition

ln(x) - ln(y) = ln(x/y)

dx[ln(x)] = lim (Δx->0) [(ln((x + Δx) / x)) / Δx]

dx[ln(x)] = lim (Δx->0) [(1 / Δx)(ln(1 + Δx/x)]

Let u = x / Δx

As Δx approaches infinite, u approaches 0.

1 / Δx = u / x

Δx / x = 1 / u

dx[ln(x)] = lim (u->infinite) [(u / x)(ln(1 + 1/u)], by substitution

dx[ln(x)] = lim (u->infinite) [(1 / x)ln(1 + 1/u)^u]

1/x is a constant, so

dx[ln(x)] = (1 / x)lim (u->infinite) [ln(1 + 1/u)^u]

From the limit of the composition of functions,...

dx[ln(x)] = (1 / x)ln(lim (u->infinite) [(1 + 1/u)^u])

lim (u->infinite) [(1 + 1/u)^u] = e, by definition

dx[ln(x)] = (1 / x)ln(e)

dx[ln(x)] = 1 / x